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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?

I tried with Tchebyshev inequality on sets ${a, b, c}$ and ${a^2, b^2 , c^2}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.

I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.

You can do it using the mean of the $m$ power for $m=frac{3}{2}$ (or convexity of $x^frac{3}{2}$ for that matter) :

We have that $frac{x^{frac{3}{2}}+y^{frac{3}{2}}+z^frac{3}{2}}{3} geq (frac{x+y+z}{3})^frac{3}{2}$ for any $x,y,z > 0$

Substituting $x=a^2, y=b^2, z=c^2$ we obtain $frac{a^3+b^3+c^3}{3} geq (frac{a^2+b^2+c^2}{3})^frac{3}{2} = 9^frac{3}{2}=27$ and therefore the sought minimum is $3cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.

From the power mean inequality we know that

$$sqrt[q]{sum_{i=1}^nw_ix_i^q} geq sqrt[p]{sum_{i=1}^nw_ix_i^p} quad forall p <q$$

where $ sum_{i=1}^n w_i=1 $

In your case $p=2, q=3$ and $n=3$. Hence,

$$left(dfrac{a^3+b^3+c^3}{3}right)^{1/3}ge left(dfrac{a^2+b^2+c^2}{3}right)^{1/2}=3$$

Okay, the question requires knowledge of the AM-GM inequality, which you do have.
You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
The answer is $81$.
Hope this helps.

For $a=b=c=3$ we obtain a value $81$.

We'll prove that it's a minimal value.

Indeed, we need to prove that $$a^3+b^3+c^3geq81,$$ which is true because
$$a^3+b^3+c^3-81=sum_{cyc}(a^3-27)=$$
$$=sum_{cyc}left(a^3-27-frac{9}{2}(a^2-9)right)=frac{1}{2}sum_{cyc}(a-3)^2(2a+3)geq0.$$
Also, Holder helps:
$$a^3+b^3+c^3=sqrt{frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}geqsqrt{frac{(a^2+b^2+c^2)^3}{3}}=81.$$